This set of Astronautics Multiple Choice Questions & Answers (MCQs) focuses on “Newton’s Law of Universal Gravitation”.

1. The force-distance relationship in Newton’s Law of Universal Gravitation is also referred to as the __________

a) inverse-exponent law

b) exponential-decay law

c) inverse-cube law

d) inverse-square law

View Answer

Explanation: According to Newton’s Law of Universal Gravitation, the gravitational force between two bodies falls with the square of the distance between them. This relationship is usually termed the ‘inverse-square’ law. For instance, doubling the distance between two objects, results in a force which is four times smaller.

2. The value of the gravitational constant does vary with ____________

a) distance

b) mass

c) distance and mass

d) density

View Answer

Explanation: The gravitational constant ‘G’ is a universal constant and remains fixed, independent of mass and/or distance.

3. The acceleration due to gravity of a body is directly proportional to ____________

a) the body’s mass

b) the gravitational constant

c) the body’s mass as well as the gravitational constant

d) its distance from a massive central object

View Answer

Explanation: We know that, according to Newton’s law, the gravitational force F experienced by an object of mass ‘m’ due to a massive body ‘M’ is given by

\(F=G\frac{Mm}{r^2}\)

where ‘r’ is the distance between the two and ‘G’ is the gravitational constant. Also, according to Newton’s second law, F=ma, where ‘a’ is the acceleration of the object. Here, ‘a’ is the same as ‘g’, namely, the acceleration due to gravity. So, F=mg.

Comparing the two equations, we have

\(g= \frac{GM}{r^2}\)

From the above equation, we see that ‘g’ is directly proportional to both ‘G’ and ‘M’.

4. How far does an Earth-based object travel vertically in the 1st second of its free fall?

a) 9.8m

b) 4.9m

c) 19.6m

d) 10m

View Answer

Explanation: The formula for distance travelled by an Earth-based object in free fall in a given amount of time ‘t’ is given by D=4.9t

^{2}. Here, t = 1, so the distance travelled would by 4.9*(12), i.e., 4.9 meters.

5. If the distance between two objects is tripled, then by how many times must each body’s mass be increased in order to prevent any change in the magnitude of the gravitational force between the two?

a) 3

b) 9

c) 1.5

d) 4.5

View Answer

Explanation: Here, originally, F = G (m

_{1}m

_{2}/r

^{2}). When the distance is tripled, the new gravitational force becomes 1/9th of the original. In order to equalize the new magnitude to the original force, each mass must increase by three times.

\(F=\frac{G(3m_1)(3m_2)}{(3r)^2}=\frac{9Gm_1m_2}{9r^2}=G\frac{m_1m_2}{r^2}\)

6. Which of the following properties does not influence the acceleration due to gravity ‘g’?

a) Mass of the central body

b) Mass of an object bound to the central body

c) Distance of an object from the central body

d) Shape of the object

View Answer

Explanation: The acceleration due to gravity is independent of how heavy an object bound to the central body is. All bodies fall with the same rate (excluding other factors like atmospheric drag, etc.) irrespective of their masses. ‘g’ depends only on the mass of the central body and how far away the object is from it.

7. The gravitational field emanating from any material object _________

a) is always infinite

b) exists only in the presence of any external body

c) is zero at an infinite distance from the source object

d) is always finite and independent of the presence of any external body

View Answer

Explanation: The gravitational field of a body is always finite and never approaches zero. At enormous distances from the object, the strength of the field is negligibly small and extremely tough to measure, but is nevertheless finite and omnipresent. Moreover, the gravity field of an object always exists whether or not there is another object nearby. The gravitational ‘force’ however requires the presence of external bodies upon which the force will be exerted.

8. The motion of the planets around the Sun can be explained by ____________

a) Newton’s law of gravitation

b) Newton’s laws of motion

c) Kepler’s laws

d) Newton’s law of gravitation along with his laws of motion

View Answer

Explanation: The movement of the heavenly bodies of our solar system around the Sun could finally be explained when Newton reconciled his law of universal gravitation with his laws of motion. In order to solve the resulting equations, the famed scientist had to invent a new branch of mathematics, which he named ‘calculus’.

9. Assuming the Moon’s mass to be 7.35 x 10^{22} kg and its radius as 1737 kilometers, the acceleration due to gravity at the Moon’s surface is ______________

a) 9.8 m/s^{2}

b) 1.62 m/s^{2}

c) 5.7 m/s^{2}

d) 4.9 m/s^{2}

View Answer

Explanation: We know that the formula for the acceleration due to gravity is

\(g= \frac{GM}{r^2}\)

Here, M = 7.35 x 10

^{22}kg and r = 1737 x 10

^{3}meters. Substituting these values in the above equation, we get g = 1.62 m/s

^{2}.

10. Assuming the mass of Saturn to be constant, if its radius were to increase, then the acceleration due to gravity at its surface would _________

a) increase

b) decrease

c) remain unchanged

d) first increase and then stay constant beyond a certain size

View Answer

Explanation: Here, ‘g’ and ‘r’ are related by the following equation:

\(g= \frac{GM}{r^2}\)

So clearly, as ‘r’ increases, the acceleration due to gravity at its surface would fall with the square of its radius.

NOTE: Generally, ‘r’ denotes the distance between the center of mass of an object and another point. Here, that point lies on the surface of the planet, so it can be considered as the radius of Saturn.

**Sanfoundry Global Education & Learning Series – Astronautics.**

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