Astronomy Quiz: Newtons Law Of Gravitation

20 questions · exam conditions

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Newtons Law Of GravitationQuestion 1 of 20

A cannon on a hypothetical, airless planet fires a projectile horizontally. As the initial launch speed is increased, the projectile travels farther before hitting the ground. According to Newton's thought experiment, what happens when the horizontal launch speed is precisely the value required for a stable, circular orbit just above the planet's surface?

The gravitational force on the projectile becomes zero, allowing it to coast in a straight line.

The projectile travels so fast that it escapes the planet's gravitational pull and moves into interstellar space.

An outward centrifugal force is generated that exactly cancels the inward pull of gravity, causing the projectile to maintain a constant altitude.

The projectile's curved trajectory matches the planet's curvature, so it continuously falls without getting closer to the surface.

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Astronomy Quiz

Astronomy Quiz: Newtons Law Of Gravitation

Practice Newtons Law Of Gravitation in Astronomy with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Newtons Law Of Gravitation, giving you a quick way to practice the rules, question types, and explanations that matter most for Astronomy.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The gravitational force on the projectile becomes zero, allowing it to coast in a straight line.
The projectile travels so fast that it escapes the planet's gravitational pull and moves into interstellar space.
An outward centrifugal force is generated that exactly cancels the inward pull of gravity, causing the projectile to maintain a constant altitude.
The projectile's curved trajectory matches the planet's curvature, so it continuously falls without getting closer to the surface. (correct answer)

Explanation: This question tests your understanding of orbital mechanics and Newton's famous thought experiment about launching a cannonball fast enough to orbit Earth. The key insight is recognizing what happens when an object moves at orbital velocity. When the projectile reaches the precise speed for circular orbit, something remarkable occurs: its trajectory curves at exactly the same rate as the planet's surface curves away beneath it. The projectile is constantly falling due to gravity, but because it's moving forward so quickly, the ground "falls away" at the same rate. This creates a perpetual free-fall situation where the projectile maintains constant altitude while continuously accelerating toward the planet's center. Choice A is incorrect because gravitational force never becomes zero—gravity is what keeps the projectile in orbit by providing the centripetal acceleration needed for circular motion. Choice B describes escape velocity, which is much faster than orbital velocity (about 1.4 times faster). At orbital speed, the projectile doesn't have enough energy to escape. Choice C uses the problematic concept of "centrifugal force." While this fictitious force appears in rotating reference frames, the real physics involves gravity providing centripetal acceleration—there's no force "canceling" gravity. Choice D correctly describes orbital motion: the projectile's parabolic trajectory matches the planet's curvature, creating continuous free-fall without surface impact. Remember this key distinction: orbital velocity creates a balance between forward motion and gravitational acceleration, not a balance of forces. The object is always accelerating toward the planet's center, but its forward speed prevents it from getting closer to the surface.

Question 2

Two spherical asteroids, P and Q, are in deep space. Asteroid P has three times the mass of asteroid Q ( $M_P = 3M_Q$ ). The radius of asteroid P is also twice the radius of asteroid Q ( $R_P = 2R_Q$ ). What is the ratio of the surface gravity of P to the surface gravity of Q ( $g_P / g_Q$ )?

3/4 (correct answer)
3/2
3
12

Explanation: Surface gravity (g) is the gravitational acceleration at the surface of an object, given by

g = GM/R^2

. The ratio of the surface gravities is

g_P/g_Q = (GM_P/R_P^2) / (GM_Q/R_Q^2) = (M_P/M_Q) \times (R_Q/R_P)^2

. Substituting the given relations

M_P = 3M_Q

and

R_P = 2R_Q

gives:

g_P/g_Q = (3M_Q/M_Q) \times (R_Q/(2R_Q))^2 = 3 \times (1/2)^2 = 3 \times 1/4 = 3/4

. B: This result is obtained if the radius is not squared (3/2). C: This is the result if the radius is ignored completely. D: This is the result if the radius ratio is squared but inverted (

3 \times 2^2 = 12

Question 3

An exoplanet, 'Planet X,' is discovered with a radius twice that of Earth ( $R_X = 2R_E$ ) and a mass eight times that of Earth ( $M_X = 8M_E$ ). An object has a weight of 700 N on the surface of Earth. What would be the object's approximate weight on the surface of Planet X?

700 N
1400 N (correct answer)
2800 N
5600 N

Explanation: Weight is the force of gravity, given by

W = GmM/R^2

. The ratio of the weight on Planet X to the weight on Earth is

W_X/W_E = (GmM_X/R_X^2) / (GmM_E/R_E^2) = (M_X/M_E) \times (R_E/R_X)^2

. We are given

M_X = 8M_E

and

R_X = 2R_E

. Substituting these values gives

W_X/W_E = (8M_E/M_E) \times (R_E/(2R_E))^2 = 8 \times (1/2)^2 = 8 \times (1/4) = 2

. So, the weight on Planet X is twice the weight on Earth:

2 \times 700\text{ N} = 1400\text{ N}

. A: This would happen if the mass-to-radius-squared ratio were the same as Earth's. C: This is the result of not squaring the radius ratio (

8/2 = 4

, so

4 \times 700 = 2800

). D: This is the result of only considering the change in mass (

8 \times 700 = 5600

Question 4

A satellite in a low Earth orbit experiences a small amount of atmospheric drag. This drag force continuously removes energy from the satellite's orbit. What is the qualitative effect of this drag on the satellite's altitude and orbital speed as it gradually spirals inward?

Its altitude decreases, and its speed decreases.
Its altitude decreases, and its speed increases. (correct answer)
Its altitude increases, and its speed decreases.
Its altitude and speed both remain constant until final reentry.

Explanation: This is a classic orbital paradox. Drag is a non-conservative force that removes total mechanical energy (kinetic + potential) from the orbit. An orbit with less energy is a lower orbit, so the satellite's altitude decreases. However, for a circular orbit, the orbital speed is given by

v = \sqrt{GM/r}

. As the orbital radius

r

decreases, the orbital speed

v

must increase. The loss in potential energy is greater than the loss in total energy, so the 'missing' energy is converted into kinetic energy, causing the satellite to speed up as it falls. A: This is the intuitive but incorrect answer, as one expects drag to slow an object down. C: This is incorrect; drag cannot add energy to raise an orbit. D: This is incorrect; the orbit is not stable and must change.

Question 5

A rocket launches a probe into a stable, low circular orbit around Earth. To send the probe to the outer solar system, the mission planners must fire the probe's engine. How must the engine firing change the probe's state to allow it to escape Earth's gravity?

It must fire in the direction of motion to increase the probe's speed to at least the escape velocity. (correct answer)
It must fire against the direction of motion to slow the probe down, causing it to fall away from the stable orbit.
It must fire perpendicular to the orbital path, pushing the probe 'outward' from Earth until gravity becomes negligible.
It must fire continuously to provide a constant thrust that directly overpowers the pull of Earth's gravity.

Explanation: To escape a gravitational field, an object must have enough kinetic energy to overcome the gravitational potential energy binding it. This corresponds to reaching a critical speed known as escape velocity. For any given altitude, escape velocity is

\sqrt{2}

times the circular orbital velocity. The most efficient way to increase kinetic energy is to apply thrust in the direction of motion (a prograde burn), increasing the probe's speed until it equals or exceeds the escape velocity at that point. B: Firing retrograde would cause the probe to lose energy and fall into a lower orbit or de-orbit entirely. C: Firing radially outward is inefficient and does not optimally increase the orbit's energy. D: A continuous burn is not necessary; escape is determined by achieving a sufficient instantaneous velocity.

Question 6

The orbit of Uranus was observed to deviate slightly from the path predicted by Newton's laws of motion and gravitation when considering only the Sun and the other known planets. In the 19th century, what was the most scientifically sound conclusion drawn from this discrepancy?

Newton's law of gravitation must be flawed and does not apply accurately at such large distances from the Sun.
The measurements of Uranus's position were systematically incorrect due to the limitations of 19th-century telescopes.
An undiscovered massive object, likely another planet, must be gravitationally perturbing Uranus's orbit. (correct answer)
A non-gravitational force, like the solar wind, becomes the dominant factor in determining orbits in the outer solar system.

Explanation: The observed perturbations in Uranus's orbit were taken as a test of Newton's law of gravitation. Rather than abandoning the law, astronomers Urbain Le Verrier and John Couch Adams trusted it and hypothesized that the gravitational pull of an unseen planet beyond Uranus was causing the deviations. Their calculations successfully predicted the location of this new planet, which was subsequently discovered and named Neptune. This was a triumphant confirmation of the predictive power of Newton's law. A: While a possibility, modifying a successful law is a last resort. The first step is to see if the law can explain the observation with new, unseen factors. B: While measurement error is always a concern, the systematic nature of the deviation pointed to a physical cause. D: Non-gravitational forces like solar wind are far too weak to cause the observed perturbations in a giant planet's orbit.

Question 7

Jupiter's mass is approximately 318 times that of Earth. Let $F_{J \to E}$ be the magnitude of the gravitational force exerted by Jupiter on Earth, and $a_E$ be the magnitude of Earth's resulting acceleration. Similarly, let $F_{E \to J}$ and $a_J$ be the force and acceleration experienced by Jupiter due to Earth. Which statement accurately compares these quantities?

$F_{J \to E} > F_{E \to J}$ and $a_E > a_J$
$F_{J \to E} = F_{E \to J}$ and $a_E > a_J$ (correct answer)
$F_{J \to E} = F_{E \to J}$ and $a_E = a_J$
$F_{J \to E} > F_{E \to J}$ and $a_E < a_J$

Explanation: According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The gravitational force Earth exerts on Jupiter is equal in magnitude and opposite in direction to the force Jupiter exerts on Earth. Therefore,

F_{J \to E} = F_{E \to J}

. According to Newton's second law of motion (

F=ma

, or

a=F/m

), acceleration is inversely proportional to mass for a given force. Since Earth has a much smaller mass than Jupiter, it will experience a much larger acceleration. Therefore,

a_E > a_J

. A: This incorrectly assumes the more massive object exerts a greater force. C: This correctly identifies the force relationship but incorrectly assumes equal accelerations. D: This incorrectly identifies both the force and acceleration relationships.

Question 8

The Moon is the primary cause of Earth's ocean tides. Which statement provides the most accurate physical explanation for why the gravitational pull of the Moon results in two high tides on opposite sides of the Earth at any given time?

The Moon's gravity pulls the ocean water on the near side towards it, while centrifugal force from Earth's rotation pushes water out on the far side.
The Moon's gravity creates a single tidal bulge, which then sloshes back and forth in the ocean basins to create a second, opposite high tide.
The Moon's gravity pulls on the oceans, and the Sun's gravity pulls on the opposite side of the Earth, creating two bulges.
The Moon's gravitational force is stronger on the side of the Earth nearer the Moon and weaker on the far side, creating a differential force that stretches the oceans. (correct answer)

Explanation: When you encounter questions about tidal mechanics, focus on understanding gravitational force as a function of distance rather than simple attraction. The key to understanding Earth's dual tidal bulges lies in recognizing that gravity weakens with distance. The Moon pulls most strongly on the side of Earth closest to it, moderately on Earth's center, and weakest on the far side. This creates a differential force that literally stretches our planet and its oceans. The near side gets pulled toward the Moon more than Earth's center does, while the far side gets pulled less than the center. This stretching effect creates two bulges: one where water is pulled toward the Moon, and another where water is "left behind" as the solid Earth gets pulled away from it. Answer D correctly describes this differential gravitational force mechanism. Answer A incorrectly attributes the far-side bulge to centrifugal force from Earth's rotation, but Earth's rotation doesn't create the tidal pattern—the Moon's varying gravitational pull does. Answer B suggests tides result from water sloshing between a single bulge, which misrepresents the simultaneous existence of two distinct bulges. Answer C incorrectly claims the Sun creates the opposite bulge, when actually both bulges result from the Moon's differential gravitational effects. Remember this key principle: tides aren't simply about the Moon "pulling" water—they're about the Moon pulling different parts of the Earth-ocean system with different strengths, creating a stretching effect that produces two simultaneous high tides on opposite sides of the planet.

Question 9

A spacecraft is in a stable circular orbit around a planet. The spacecraft briefly fires its thrusters in the direction of its motion (a prograde burn). Qualitatively, what is the shape of the new, resulting orbit immediately after the burn is complete?

A larger, stable circular orbit that is concentric with the first.
A spiral path that moves the spacecraft steadily away from the planet.
An elliptical orbit, with the point of the thruster firing being the new perigee. (correct answer)
An elliptical orbit, with the point of the thruster firing being the new apogee.

Explanation: A prograde burn increases the spacecraft's kinetic energy and velocity at that specific point in its orbit. The spacecraft now has too much velocity for a circular orbit at that radius. Its new total energy is higher, corresponding to a larger orbit. The point where the burn occurred is the only point shared by the old and new orbits, and it becomes the point of closest approach (perigee) of the new, larger elliptical orbit. The spacecraft will then travel out to a new, higher farthest point (apogee). A: To achieve a larger circular orbit, a second burn is required at the apogee of the intermediate elliptical transfer orbit. B: A spiral path would require continuous thrust. D: This would be the result of a retrograde (braking) burn, which would lower the orbit.

Question 10

An astronaut aboard the International Space Station (ISS) is said to be 'weightless.' The ISS orbits at an altitude where the gravitational force from Earth is approximately 90% of its strength at the surface. Which statement best explains this phenomenon?

The ISS is so far from Earth that the gravitational force is negligible, allowing the astronaut to float freely.
The outward-pushing centrifugal force generated by the orbit perfectly balances the inward pull of gravity, creating a zero-g environment.
The astronaut and the ISS are both in a constant state of free fall around the Earth, accelerating at nearly the same rate. (correct answer)
The high orbital speed of the ISS generates an aerodynamic lift that counteracts gravity, similar to an airplane in flight.

Explanation: The feeling of weightlessness in orbit is not due to a lack of gravity, but to being in a constant state of free fall. The ISS and everything in it are continuously falling towards Earth due to gravity. However, they also have a high tangential velocity, so as they fall, the Earth's surface curves away beneath them. Since the astronaut and the station are accelerating together, the astronaut does not press against the floor, creating the sensation of weightlessness. A: This is incorrect; gravity at the ISS's altitude is still about 90% of surface gravity. B: This invokes the concept of 'centrifugal force,' which is a fictitious force in an accelerating reference frame. The more fundamental explanation is that gravity provides the required centripetal force for the circular motion. D: This is incorrect; there is virtually no air in orbit to provide aerodynamic lift.

Question 11

Consider a hypothetical universe where the universal gravitational constant, $G$ , is twice as large as in our universe. If a planet in that universe has the same mass and orbital radius as Earth, how would its orbital period compare to Earth's current period of one year?

The period would be twice as long.
The period would be half as long.
The period would be shorter by a factor of $\sqrt{2}$ . (correct answer)
The period would be longer by a factor of $\sqrt{2}$ .

Explanation: For a circular orbit, the gravitational force provides the centripetal force:

GMm/r^2 = mv^2/r

. The orbital period

T

2\pi r/v

. From the force equation,

v = \sqrt{GM/r}

. Substituting this into the period equation gives

T = 2\pi r / \sqrt{GM/r} = 2\pi \sqrt{r^3/GM}

. This shows that

T \propto 1/\sqrt{G}

. If

G

is doubled (

G' = 2G

), the new period

T'

will be

T' \propto 1/\sqrt{2G} = (1/\sqrt{2})(1/\sqrt{G})

. Therefore, the new period would be the original period divided by

\sqrt{2}

. A: This incorrectly assumes

T \propto G

. B: This incorrectly assumes

T \propto 1/G

, forgetting the square root. D: This inverts the relationship.

Question 12

Astronomers observe that stars in the outer regions of spiral galaxies orbit the galactic center much faster than predicted by Newton's law of gravitation based on the visible matter (stars, gas, and dust). If Newton's law is assumed to be correct, what does this observation directly imply about the composition of these galaxies?

The gravitational constant G must be larger on galactic scales than it is in the solar system.
The visible matter in the outer regions of galaxies must be significantly more massive than it appears.
There is an error in Newton's law of gravitation, which breaks down at very low accelerations.
The galaxies must contain a large amount of non-luminous 'dark matter' which provides the additional gravitational force. (correct answer)

Explanation: When you encounter questions about galactic rotation curves, you're dealing with one of astronomy's most significant discoveries that led to our understanding of dark matter. The key principle here is that if Newton's laws are correct, orbital velocities should decrease with distance from the galactic center, just like planets farther from the Sun orbit more slowly. The observation shows stars moving faster than expected in outer galactic regions. If we assume Newton's gravitational law is correct, this excess velocity can only be explained by additional mass providing extra gravitational force. Since this mass isn't visible as stars, gas, or dust, it must be "dark matter" – matter that doesn't emit or absorb light but does exert gravitational influence. This makes D correct. A is wrong because changing the gravitational constant G would affect all scales equally, not just galactic ones, and would contradict countless other observations. B incorrectly suggests the visible matter is more massive than it appears – but the problem is that there isn't enough visible matter, not that we're underestimating its mass. C proposes modifying Newton's laws, but the question specifically asks what the observation implies if Newton's law is assumed correct. Study tip: Remember that dark matter questions often test whether you can distinguish between modifying physical laws versus adding unseen matter. When a problem states "assume the law is correct," you're being guided toward the dark matter explanation rather than modified gravity theories.

Question 13

A communications satellite is in a stable circular orbit at an altitude of one Earth radius ( $R_E$ ) above the Earth's surface. To move it to a different stable circular orbit, its altitude is increased to 5.5 $R_E$ above the surface. By approximately what factor does the gravitational force exerted by the Earth on the satellite change?

It decreases by a factor of 3.25.
It decreases by a factor of 5.5.
It decreases by a factor of 10.6. (correct answer)
It decreases by a factor of 30.25.

Explanation: Newton's law of gravitation states that force is inversely proportional to the square of the distance between the centers of the objects (

F \propto 1/r^2

). The distance must be measured from the center of the Earth. The initial distance is

r_1 = R_E + 1 R_E = 2 R_E

. The final distance is

r_2 = R_E + 5.5 R_E = 6.5 R_E

. The ratio of the final force to the initial force is

F_2/F_1 = (r_1/r_2)^2 = (2 R_E / 6.5 R_E)^2 = (2/6.5)^2 \approx (0.308)^2 \approx 0.0947

. The force decreases by a factor of

1/0.0947 \approx 10.56

. The closest answer is 10.6. A: This is the ratio of the distances (6.5/2), a common mistake where the inverse square relationship is forgotten. B: This uses the change in altitude (5.5) directly, ignoring the initial altitude and the Earth's radius. D: This results from squaring the ratio of the altitudes (

5.5^2

), neglecting to add the Earth's radius to find the orbital distance from the center.

Question 14

Two satellites, Star-A and Star-B, are in stable, circular orbits around the same planet. Star-A has a mass of 1,000 kg and orbits at a radius of $R$ . Star-B has a mass of 2,000 kg and orbits at a radius of $4R$ . What is the ratio of the orbital speed of Star-A to the orbital speed of Star-B ( $v_A / v_B$ )?

1/2
1
2 (correct answer)
4

Explanation: The speed of a satellite in a circular orbit is given by the formula

v = \sqrt{GM/r}

, where

M

is the mass of the central body (the planet) and

r

is the orbital radius. The mass of the satellite itself does not affect its orbital speed. So,

v_A = \sqrt{GM/R}

and

v_B = \sqrt{GM/(4R)} = (1/2)\sqrt{GM/R}

. The ratio is

v_A / v_B = (\sqrt{GM/R}) / ((1/2)\sqrt{GM/R}) = 2

. A: This is the inverse of the correct ratio. B: This would be correct if the orbital radii were the same. D: This would be the result if speed were inversely proportional to the radius (

v \propto 1/r

) instead of the square root of the radius (

v \propto 1/\sqrt{r}

Question 15

$F_{J \to E} > F_{E \to J}$ and $a_E > a_J$
$F_{J \to E} = F_{E \to J}$ and $a_E > a_J$ (correct answer)
$F_{J \to E} = F_{E \to J}$ and $a_E = a_J$
$F_{J \to E} > F_{E \to J}$ and $a_E < a_J$

F_{J \to E} = F_{E \to J}

. According to Newton's second law of motion (

F=ma

, or

a=F/m

), acceleration is inversely proportional to mass for a given force. Since Earth has a much smaller mass than Jupiter, it will experience a much larger acceleration. Therefore,

a_E > a_J

Question 16

A comet travels in a highly elliptical orbit around the Sun. Let Point P be its perihelion (closest approach to the Sun) and Point A be its aphelion (farthest point from the Sun). Which of the following correctly describes the comet's speed and the Sun's gravitational force upon it at these two points?

The force is greatest at P, and the comet's speed is greatest at P. (correct answer)
The force is greatest at A, and the comet's speed is greatest at A.
The force is the same at both P and A, but the speed is greatest at P.
The force is greatest at P, but the speed is greatest at A.

Explanation: According to Newton's law of gravitation, the gravitational force is inversely proportional to the square of the distance (

F \propto 1/r^2

). Therefore, the force is greatest at perihelion (P), the point of closest approach. Due to the conservation of angular momentum, the comet must speed up as it gets closer to the Sun and slow down as it moves farther away. Thus, its speed is also greatest at perihelion. B: This reverses both relationships. C: This incorrectly states the force is constant. D: This incorrectly relates speed and distance.

Question 17

Its altitude decreases, and its speed decreases.
Its altitude decreases, and its speed increases. (correct answer)
Its altitude increases, and its speed decreases.
Its altitude and speed both remain constant until final reentry.

v = \sqrt{GM/r}

. As the orbital radius

r

decreases, the orbital speed

v

Question 18

Imagine a spacecraft traveling along a straight line from the Earth's center to the Moon's center. At what point along this path is the net gravitational force exerted on the spacecraft by the Earth and Moon equal to zero? Earth's mass is approximately 81 times the Moon's mass.

At the point exactly halfway between the Earth and the Moon.
At a point 9/10 of the way from the Earth to the Moon. (correct answer)
At a point 80/81 of the way from the Earth to the Moon.
At a point closer to the Earth than to the Moon.

Explanation: For the net force to be zero, the magnitudes of the gravitational forces from the Earth and Moon must be equal:

F_E = F_M

. Using Newton's law of gravitation,

GM_Em/r_E^2 = GM_Mm/r_M^2

, where

r_E

and

r_M

are the distances to the spacecraft from the centers of the Earth and Moon, respectively. This simplifies to

M_E/M_M = (r_E/r_M)^2

. Given

M_E/M_M = 81

, we have

81 = (r_E/r_M)^2

, which means

r_E/r_M = 9

. The point of zero net force is 9 times farther from the Earth than from the Moon. If the total distance is D, then

r_E + r_M = D

and

r_E = 9r_M

. Solving gives

10r_M = D

, so

r_M = D/10

and

r_E = 9D/10

. The point is 9/10 of the way from Earth to the Moon. A: This ignores the large mass difference. C: This would be true if force was proportional to

1/r

, not

1/r^2

. D: This is qualitatively incorrect; the zero-force point must be closer to the less massive body (the Moon).

Question 19

700 N
1400 N (correct answer)
2800 N
5600 N

Explanation: Weight is the force of gravity, given by

W = GmM/R^2

. The ratio of the weight on Planet X to the weight on Earth is

W_X/W_E = (GmM_X/R_X^2) / (GmM_E/R_E^2) = (M_X/M_E) \times (R_E/R_X)^2

. We are given

M_X = 8M_E

and

R_X = 2R_E

. Substituting these values gives

W_X/W_E = (8M_E/M_E) \times (R_E/(2R_E))^2 = 8 \times (1/2)^2 = 8 \times (1/4) = 2

. So, the weight on Planet X is twice the weight on Earth:

2 \times 700\text{ N} = 1400\text{ N}

. A: This would happen if the mass-to-radius-squared ratio were the same as Earth's. C: This is the result of not squaring the radius ratio (

8/2 = 4

, so

4 \times 700 = 2800

). D: This is the result of only considering the change in mass (

8 \times 700 = 5600

Question 20

The gravitational force on the projectile becomes zero, allowing it to coast in a straight line.
The projectile travels so fast that it escapes the planet's gravitational pull and moves into interstellar space.
An outward centrifugal force is generated that exactly cancels the inward pull of gravity, causing the projectile to maintain a constant altitude.
The projectile's curved trajectory matches the planet's curvature, so it continuously falls without getting closer to the surface. (correct answer)