AP Physics B & C - Multiple Choice Practice Questions on Gravitation

The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:

(1) The velocity of escape from the earth’s surface is √(2GM/R) and the work done by earth’s gravitational force to bring unit mass from infinity to a height h from the earth’s surface is GM/(R+h) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass m is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.

(a) R/2

(b) R

(c) 2R

(d) 8R

(e) ∞

The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].

(2) If the body in the above problem is projected up with 20 % of the escape velocity, how high will it rise?

(a) R/24

(b) R/20

(c) R/16

(d) R/8

(e) R/4

The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]² = 0.04 GMm/R.

The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)

[Note that the gravitational potential energy is negative]

Therefore, we have

0.04 GMm/R = GMm/R – GMm/(R+h)

Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]

This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.

(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6R where R is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5R above the surface of the earth will be approximately

(a) 16 hour

(b) 12√2 hour

(c) 8√2 hour

(d) 6√2 hour

(e) 6 hour

According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have

24² α (7R)³ since the orbital radius is 6R+R = 7R and the orbital period is 24 hour.

In the case of the research satellite the period T is related to its orbital radius by

T² α (3.5R)³ since its orbital radius is 2.5R+R = 3.5R

From the above relations (on dividing),

24/T = 2^3/2

Therefore, T = 24/2^3/2 = 24/(2√2) = 12/√2 = 6√2 hour.

The following questions are meant solely for AP Physics C aspirants:

(4) Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by

(a) 1%

(b) 1.5%

(c) 2%

(d) 2.5%

(e) 3%

We have from Kepler’s third law, T² α R³ where T and R are the orbital period and orbital radius respectively.

Therefore, T = kR^3/2 where k is the constant of proportionality.

Taking logarithms, ln T = ln k + (3/2) ln R

Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.

In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].

(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will

(a) remain unchanged

(b) decrease by 2%

(c) increase by 2%

(d) increase by 1%

(e) decrease by 1%

The orbital speed v of a satellite moving around the earth in an orbit of radius r is given by

v = √(GM/r) where G is the gravitational constant and M is the mass of the earth.

[You can easily obtain this by equating the centripetal force to the gravitational pull: mv²/r = GMm/r² where m is the mass of the satellite].

This equation shows that the orbital speed will decrease when the orbital radius is increased.

Since G and M are constants, we have

dv/v = – ½ dr/r

[You will get this by taking logarithms and by differentiating, as we did in question No.4].

This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the decrease in the orbital speed because of the increase in orbital radius.

Since the orbital radius of the satellite is increased by 2%, the orbital speed is decreased by 1% [Option (e)].

(6) A communication satellite of mass m is placed initially in a temporary orbit of radius r₁ around the earth_. How much work is to be done to shift it from this orbit to a permanent orbit of greater radius r₂? (Acceleration due to gravity at the sea level = g. Radius of the earth = R)

(a) mgR²(r₂–r₁)/ r₂r₁

(b) mgR (r₂–r₁)/ 2r₂

(c) mgR²(r₂–r₁)/ 2r₂r₁

(d) mgR²(r₂–r₁)/ 2r₂r₁

(e) 2mgR²(r₂–r₁)/ r₂r₁

Total energy (K.E. + P.E.) of a satellite of mass m in an orbit of radius r is – GMm/2r where M is the mass of the earth.

Work (W) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.

Therefore, W = – GMm/2r₂ – (– GMm/2r₁)

[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].

Thus W = (GMm/2)(1/r₁ – 1/r₂) = (GMm/2)[(r₂–r₁)/ r₂r₁]

Since g = GM/R² (as you can obtain from mg = GMm/R²), we have

W = mgR²(r₂–r₁)/ 2r₂r₁

AP Physics B & C - Multiple Choice Practice Questions on Gravitation

The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:

(1) The velocity of escape from the earth’s surface is √(2GM/R) and the work done by earth’s gravitational force to bring unit mass from infinity to a height h from the earth’s surface is GM/(R+h) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass m is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.

(a) R/2

(b) R

(c) 2R

(d) 8R

(e) ∞

The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].

(2) If the body in the above problem is projected up with 20 % of the escape velocity, how high will it rise?

(a) R/24

(b) R/20

(c) R/16

(d) R/8

(e) R/4

The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]² = 0.04 GMm/R.

The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)

[Note that the gravitational potential energy is negative]

Therefore, we have

0.04 GMm/R = GMm/R – GMm/(R+h)

Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]

This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.

(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6R where R is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5R above the surface of the earth will be approximately

(a) 16 hour

(b) 12√2 hour

(c) 8√2 hour

(d) 6√2 hour

(e) 6 hour

According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have

24² α (7R)³ since the orbital radius is 6R+R = 7R and the orbital period is 24 hour.

In the case of the research satellite the period T is related to its orbital radius by

T² α (3.5R)³ since its orbital radius is 2.5R+R = 3.5R

From the above relations (on dividing),

24/T = 2^3/2

Therefore, T = 24/2^3/2 = 24/(2√2) = 12/√2 = 6√2 hour.

The following questions are meant solely for AP Physics C aspirants:

(4) Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by

(a) 1%

(b) 1.5%

(c) 2%

(d) 2.5%

(e) 3%

We have from Kepler’s third law, T² α R³ where T and R are the orbital period and orbital radius respectively.

Therefore, T = kR^3/2 where k is the constant of proportionality.

Taking logarithms, ln T = ln k + (3/2) ln R

Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.

In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].

(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will

(a) remain unchanged

(b) decrease by 2%

(c) increase by 2%

(d) increase by 1%

(e) decrease by 1%

The orbital speed v of a satellite moving around the earth in an orbit of radius r is given by

v = √(GM/r) where G is the gravitational constant and M is the mass of the earth.

[You can easily obtain by equating the centripetal force to the gravitational pull: mv²/r = GMm/r² where m is the mass of the satellite].

This equation shows that the orbital speed will decrease when the orbital radius is increased.

Since G and M are constants, we have

dv/v = – ½ dr/r

[You will get this by taking logarithms and by differentiating, as we did in question No.4].

This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the decrease in the orbital speed because of the increase in orbital radius.

Since the orbital radius of a the satellite is increased by 2%, the orbital speed is decreased by 1% [Option (e)].

(6) A communication satellite of mass m is placed initially in a temporary orbit of radius r₁ around the earth_. How much work is to be done to shift it from this orbit to a permanent orbit of greater radius r₂? (Acceleration due to gravity at the sea level = g. Radius of the earth = R)

(a) mgR²(r₂–r₁)/ r₂r₁

(b) mgR (r₂–r₁)/ 2r₂

(c) mgR²(r₂–r₁)/ 2r₂r₁

(d) mgR²(r₂–r₁)/ 2r₂r₁

(e) 2mgR²(r₂–r₁)/ r₂r₁

Total energy (K.E. + P.E.) of a satellite of mass m in an orbit of radius r is – GMm/2r where M is the mass of the earth.

Work (W) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.

Therefore, W = – GMm/2r₂ – (– GMm/2r₁)

[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].

Thus W = (GMm/2)(1/r₁ – 1/r₂) = (GMm/2)[(r₂–r₁)/ r₂r₁]

Since g = GM/R² (as you can obtain from mg = GMm/R²), we have

W = mgR²(r₂–r₁)/ 2r₂r₁

AP Physics B & C – Multiple Choice Questions (MCQ) for Practice on Gravitation

Equations to be remembered in respect of the section on gravitation were given in the post dated 9^th May 2008. A few multiple choice practice questions were discussed in the post dated 12^th May 2008 followed by a free response practice question on gravitation in the post dated 15^th May 2008. You can access all posts on gravitation on this site by clicking on the label ‘gravitation’ below this post.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are meant for AP Physics B as well as C:

(1) The speed of an artificial satellite moving in an orbit of radius r around the earth is increased by 41.4%. The satellite will

(a) continue to move in the same orbit

(b) move in an orbit of radius 1.414 r

(c) move in an orbit of radius 0.414 r

(d) move in an orbit of radius 0.586 r

(e) escape into the outer space

The orbital speed v of any satellite is given by

v =√(gr) where g is the acceleration due to gravity at the orbit of radius r.

[You will get this by equating the centripetal force to the gravitational pull:

mv²/r = mg]

The escape velocity (v_escape) of a body at a point distant r from the centre of the earth where the acceleration due to gravity is equal to g is given by

v_escape =√(2gr)

Thus the escape velocity is √2 times the orbital speed.

The speed v of the satellite becomes 1.414 times the initial value on increasing the speed by 41.4%. Now, 1.414 v = √2 v. Therefore the new speed is the escape speed so that the satellite will escape into the outer space.

(2) A launch vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. What is the minimum energy to be spent by the launch vehicle on the satellite so that the satellite will move in a circular orbit of radius 7R? (Gravitational constant = G)

(a) GMm /7R

(b) 6GMm /7R

(c) 6GMm /14R

(d) 13GMm /14R

(e) GMm /14R

The satellite of mass m is initially at rest and therefore its initial energy (E_i) is its gravitational potential energy given by

E_i = – GMm/R

When the satellite is in its orbit of radius 7R, its final energy (E_f) is given by

E_f = – GMm/(2×7R) = – GMm/(14R)

[Note that the total energy of a satellite in an orbit of radius r is – GMm/2r]

The minimum energy required for placing the satellite in its orbit is E_f – E_i given by

E_f – E_i = – [GMm/(14R)] – [– GMm/R]

= (GMm/R) – [GMm/(14R)]

= 13GMm /14R

(3) A small body is projected vertically up with a speed equal to half the escape speed from the earth’s surface. If the radius of the earth is R, what is the maximum height (h) reached by the body? (Neglect air resistance).

(a) R/3

(b) R/2

(c) R

(d) 2R

(e) 5R

When the body projected vertically reaches the maximum height, its speed will be zero since its initial kinetic energy us used up in increasing its gravitational potential energy. Therefore we have

½ m (v_e/2)² = [–GMm /(R+h)] – [–GMm /R] where m is the mass of the body, M is the mass of the earth, v_e is the escape speed from the earth’s surface and G is the gravitational constant.

[The first term on the right hand side of the above equation is the gravitational potential energy of the body at height h. The second term is the gravitational potential energy of the body on the surface of the earth].

Since v_e = √(2GM/R), the above equation becomes

GM /4R = (GM /R) –[GM /(R+h)]

Or, GM /4R = GMh /[R(R+h)

Therefore, ¼ = h/(R+h) from which h = R/3

(4) The orbit of a planet moving around the sun is elliptical with the sun S at one focus of the ellipse. If the orbital speeds of the planet while at points A and B (Fig.) are V₁ and V₂ and the distances SA and SB are d₁ and d₂ respectively, V₁/ V₂ is equal to

(a) √(d₁ /d₂)

(b) √(d₂ /d₁)

(c) d₁ /d₂

(d) d₂ /d₁

(e) (d₂ + d₁) /(d₂ – d₁)

To obtain the answer you may apply Kepler’s law of areas (in respect of plnetary motion), which says that the straight line joining the sun to the planet sweeps equal areas in equal intervals of time. The adjoining figure shows path lengths CD and EF traced by the planet in equal intervals (say, ∆t) and hence the areas CADS and EBFS must be equal. If ∆t is sufficiently small, CD and EF will become straight line segments and the areas will become triangular. Equating the areas of these triangles, we have

½ ×CD×AS = ½ ×EF×BS

Or, ½ × V₁ ∆t × d₁ = ½ × V₂ ∆t × d₂

Therefore, V₁/ V₂ = d₂ /d₁

[You can easily get the answer from angular momentum conservation:

m V₁ d₁ = m V₂ d₂

where m is the mass of the planet. This gives V₁/ V₂ = d₂ /d₁]

(5) We know that the gravitational force between two point masses is inversely proportional to the square of the distance between them. If the gravitational force between two point masses were (let us imagine) inversely proportional to the n^th power of the distance between them, what would be the relation between the orbital period (T) of a planet around the sun and the mean distance (r) of the planet from the sun?

(a) T² α r²ⁿ

(b) T² α rⁿ

(c) T² α r²ⁿ⁺¹

(d) T² α r⁽ⁿ⁺¹⁾

(e) T² α r^{(n – 1)}

For simplicity, let us assume that the orbit is circular. (Circular orbit is only a special case of the general elliptical orbit and the relation between T and r will be unchanged). The centripetal force required for the circular motion is supplied by the gravitational pull so that we have

mrω² = GMm/rⁿ where m is the mass of the planet, ω is its orbital angular velocity, M is the mass of the sun and G is the gravitational constant.

The above equation gives ω² = GM/r⁽ⁿ⁺¹⁾

Since ω = T/2π, we have

(T/2π)² = GM/r⁽ⁿ⁺¹⁾

Therefore, T² α r⁽ⁿ⁺¹⁾

You will find some useful multiple choice questions on gravitation (with solution) at physicsplus.