YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONSstage2physicsahs.weebly.com/uploads/4/6/4/2/46424629/gravitation... · than the planet, M. Assume that the satellite moves around the

YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONS Name:

QUESTION 1 (1995 EXAM)

(a) State Newton’s Universal Law of Gravitation in words

(3 marks)

(b) A satellite of mass (m) moves in orbit of a planet with mass (M). The satellite, m, is smaller in mass

than the planet, M. Assume that the satellite moves around the planet in a circular orbit with a

radius, R with a constant speed, v.

(i) Explain how it is possible for the gravitational force to cause the satellite to accelerate while

its speed remains constant.

(2 marks)

Between any two masses, there exists a mutual attractive force.

This force is directly proportional to the product of the masses and inversely

proportional to the square of the distance between them.

This force acts along a line between the centres of the masses.

Its speed maybe constant in magnitude but the direction of this motion is

constantly changing over time due to the gravitational force constantly

pulling the satellite towards the central mass. Since this force is constant

and the speed is constantly changing per unit time, the body is acceleration.

(ii) If T is the period of the satellite in its orbit around the planet, show that the radius of the

orbit of the satellite is r = √𝐺𝑚2𝑇2

4𝜋2

3 given that T2 =

4𝜋2𝑟3

Gm2

(2 marks)

(iii) Calculate the moon’s orbital radius given that its period of rotation is approximately 27 days

and 7 hours (2.3582 x 106 s)

(3 marks)

𝐓𝟐 = 𝟒𝛑𝟐𝐫𝟑

𝐆𝐦𝟐

𝐫𝟑 = 𝐆𝐦𝟐𝐓𝟐

𝟒𝛑𝟐

r = √𝐆𝐦𝟐𝐓𝟐

𝟒𝛑𝟐

𝟑


𝟒𝛑𝟐

𝟑

r = √𝟔.𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝒙 𝟓.𝟗𝟕𝟕 𝐱 𝟏𝟎𝟐𝟒 𝒙 (𝟐.𝟑𝟓𝟖𝟐 𝐱 𝟏𝟎𝟔)𝟐

𝟒𝛑𝟐

𝟑

r = √𝟐.𝟐𝟏𝟕 𝐱 𝟏𝟎𝟐𝟕

𝟒𝛑𝟐

𝟑

r = 𝟑. 𝟖𝟑 𝐱 𝟏𝟎𝟖 𝐦


Whilst orbiting the Earth, the space shuttle Endeavour had a velocity of 7.8 x 103 ms-1

(a) Calculate the radius of its circular orbit

(2 marks)

(b) Two isolated masses M and m are separated by a distance, r. The mass M is twice the mass of the

smaller body, m.

On the diagram above, draw vector arrows to illustrate the gravitational force (F) on each mass

(2 marks)

HINT: Despite one mass being larger, the gravitational force acting on each object is equal

in magnitude but opposite in direction.

𝐕 = √𝐆𝐌

𝐫 ∴ 𝐫 =

𝐆𝐌

𝐕𝟐

𝐫 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒

(𝟕. 𝟖 𝐱 𝟏𝟎𝟑)𝟐

𝐫 = 𝟔. 𝟓𝟒 𝐱 𝟏𝟎𝟔 𝐦


The uniform circular motion of a space vehicle in a circular orbit round a planet is caused by the

gravitational force between the planet and the vehicle.

(a) Calculate the magnitude and direction of the gravitational force on a space vehicle of mass

m= 1.00 x 103 kg at a distance r of 1.65 x 107 m from the centre of a planet of mass

M = 2.00 x 1025 kg given the Universal Gravitation law (𝐹 =𝐺𝑚1𝑚2

𝑟2 ).

(3 marks)

(b) Show that the speed v of the space vehicle is given by the formula 𝑣 = √𝐺𝑚2

𝑟

(2 marks)

(c) Using the relationship for the speed of the space vehicle given in part (ii) and an expression for the

period T, showing that (T2 = 4𝜋2𝑟3

Gm2)

(3 marks)

𝐅 =𝐆𝐦𝟏𝐦𝟐

𝐫𝟐

𝐅 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟏. 𝟎𝟎 𝐱 𝟏𝟎𝟑 𝐱 𝟐. 𝟎𝟎 𝐱 𝟏𝟎𝟐𝟓

(𝟏. 𝟔𝟓 𝐱 𝟏𝟎𝟕)𝟐

𝐅 = 𝟒. 𝟗𝟎 𝐱 𝟏𝟎𝟑 𝐍


𝐫𝟐 𝐅 = 𝐦𝐯𝟐

𝐫

𝐆𝐦𝟐

𝐫

𝐯𝟐

(Cancel out like terms)

𝐯 = √𝐆𝐦𝟐

𝐫

𝐯 = 𝟐𝛑𝐫

𝐓 𝐯𝟐 =

𝐆𝐦𝟐

𝐫

Squaring everything in equation one yields:

𝐯𝟐 = 𝟐𝟐𝛑𝟐𝐫𝟐

𝐓𝟐 𝐯𝟐 = 𝐆𝐦𝟐

𝐫

Cancelling 𝐯𝟐 𝐚𝐧𝐝 𝐫𝐞𝐚𝐫𝐫𝐚𝐧𝐠𝐢𝐧𝐠 𝐭𝐡𝐞 𝐮𝐧𝐢𝐭𝐬 𝐨𝐟 𝒓 gives

𝟒𝛑𝟐𝐫𝟑

𝐓𝟐 𝐆𝐦𝟐

Rearranging this gives 𝐓𝟐 = 𝟒𝛑𝟐𝐫𝟑

𝐆𝐦𝟐


(a) Calculate the magnitude of the gravitational force F on mass m = 20.0 kg, positioned at 1.00 x 106 m

above the Earth's surface. The mass of the Earth ME = 5.98 x 1024 kg and the radius of the Earth

RE = 6.38 x 106 m.

(3 marks)

(b) A space vehicle of mass m is moving at a constant speed v in a circular orbit of radius r round the

Earth.

(i) Derive an expression for v in terms of the radius r and the mass of the Earth ME.

(3 marks)

(ii) Explain why the space vehicle does not need to use rocket engines to maintain its uniform

circular motion. Ignore air resistance.

(2 marks)

r = 1.00 x 106 + 6.38 x 106 = 7.38 x 106


𝐫𝟐

𝐅 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟐𝟎 𝐱 𝟓. 𝟗𝟖 𝐱 𝟏𝟎𝟐𝟒

(𝟕. 𝟑𝟖 𝐱 𝟏𝟎𝟔)𝟐

𝐅 = 𝟏𝟒𝟔 𝐍

𝐅 =𝐆𝐦𝐌𝐄


𝐫

𝐆𝐌𝐄

𝐫

𝐯𝟐


𝐯 = √𝐆𝐌𝐄

𝐫

As the gravitation al force is supplying the centripetal acceleration and this

force is constant.


(a) State, in words, Newton's law of universal gravitation.

(2 marks)

(b) Two masses exert a force of magnitude F on each other when placed a distance d apart.

State the magnitude of the force, in terms of F, if the distance d between the masses is doubled.

(1 mark)

(c) Explain why the gravitational forces between two particles of different mass, M and m, as shown

below, are consistent with Newton's third law.

(2 marks)

Between any two masses, there exists a mutual attractive force.

This force is directly proportional to the product of the masses and inversely

proportional to the square of the distance between them.

This force acts along a line between the centres of the masses.

𝐅 ∝𝟏

𝐫𝟐 if the distance were doubled (x 2), the magnitude of F would be 4 times lower in

magnitude ( 𝟏

𝟒 its original magnitude)

The forces are equal in magnitude and opposite in direction. M exerts a force on

m and hence, m exerts the same amount of force on M, just in the opposite

direction. This is consistent with Newton’s 3rd law

(c) Using Newton's second law and law of universal gravitation, derive the expression (𝑔 =𝐺𝑚2

𝑟2 ) for the

gravitational acceleration at the Moon's surface. (M is the mass of the Moon, r the radius

of the Moon.)

(2 marks)

(d) Calculate the value of the acceleration due to gravity at the Moon's surface. (The mass of the Moon

M = 7.35 x 1022 kg and the radius of the Moon r = 1.74 x 106 m.)

(2 marks)

(e) The orbit of a geostationary satellite round the Earth is shown in the diagram below:

(i) Explain why this satellite must orbit in the same direction as the Earth rotates.

(1 mark)


𝐫𝟐 𝐅 = 𝐦𝐠

Cancel out like terms (F and m)

𝐆𝐦𝟐

𝐫𝟐 = 𝐠

𝐠 =𝐆𝐦𝟐

𝐫𝟐

𝐠 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟕. 𝟑𝟓 𝐱 𝟏𝟎𝟐𝟐

(𝟏. 𝟕𝟒 𝐱 𝟏𝟎𝟔)𝟐

𝐠 = 𝟏. 𝟔𝟐 𝐦𝐬−𝟐

In order to stay above the same location above the Earth, the satellite must

orbit in the same direction as the Earth is rotating

(ii) Explain why the orbit of this satellite must be equatorial.

(2 marks)

(f) Explain why low-altitude polar orbits are used for surveillance satellites.

(3 marks)

(g) A space vehicle is moving at constant speed in a circular orbit A round a planet, as shown in the

diagram below:

Draw and label vectors on the diagram to represent the velocity v and acceleration a of the space vehicle

at position P.

(2 marks)

The centre of the satellites’ orbit must be the centre of the earth as gravity is

providing the centripetal acceleration. Equatorial orbits satisfy this condition

and satellites in these orbits rotate with the same period as the Earth which

allows the gravitational force to act in the plane of the equator.

High resolution images are available at lower altitudes.

The polar orbits at lower altitudes allow for a greater number of successful

revolutions of the Earth such that over 24 hours, the satellite has photographed

a large % of the Earth’s surface.

Lower orbit allows for less interference from atmospheric gases and sunlight.

v

a


(a) Calculate the magnitude of the gravitational acceleration g at the Earth's surface, using Newton's

second law and the law of universal gravitation. The mass of the Earth M is 5.98 x 1024 kg and the

radius of the Earth R is 6.38 x 106 m.

(3 marks)

(b) A satellite orbits the Earth with constant speed in a circle whose radius is twice the radius of the

Earth, as shown in the diagram below. The mass of the Earth is M= 5.98 x 1024 kg and its mean

radius is R = 6.38 x 106 m.

(i) Show that the speed at which the satellite is moving is approximately 6 x 103 ms-1.

(2 marks)

𝐠 =𝐆𝐦𝟐

𝐫𝟐

𝐠 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟗𝟖 𝐱 𝟏𝟎𝟐𝟒

(𝟔. 𝟑𝟖 𝐱 𝟏𝟎𝟔)𝟐

𝐠 = 𝟗. 𝟖𝟎 𝐦𝐬−𝟐

𝐯 = √𝐆𝐦𝟐

𝐫

= √𝟔.𝟔𝟕𝐱 𝟏𝟎−𝟏𝟏 𝒙 𝟓.𝟗𝟖 𝒙 𝟏𝟎𝟐𝟒

(𝟐 𝐱 𝟔.𝟑𝟖 𝐱 𝟏𝟎𝟔)

𝐯 = 𝟓𝟕𝟕𝟒. 𝟗 𝐦𝐬−𝟏

Rounding up to 1 sig. fig gives = 6 x 103 ms-1

(ii) Calculate the period of the satellite.

(2 marks)

(iii) Explain why the satellite travels with uniform circular motion in a fixed orbit.

(3 marks)

Unlike planets which have elliptical orbits, human-made satellites have

circular orbits and therefore circular motion formulae may be used:

𝑻 = 𝟐𝝅𝒓

𝒗

𝑻 = 𝟐𝝅(𝟐 𝒙 𝟔.𝟑𝟖 𝒙 𝟏𝟎𝟔)

𝟓𝟕𝟕𝟒.𝟗 = 6.94 x 103 seconds

The gravitational force between the centres of the satellite and the larger

mass is supplying the centripetal acceleration of the satellite.

This is the ONLY force acting on the satellite and therefore according to

Newton’s 1st law, where the forces on a body are constant, the speed of

the body remains constant.

A body moving with a constant speed in a circular path is described as

having uniform circular motion.


(a) Show that the speed v of a satellite moving in an orbit of radius r round a planet of mass M is given

by, (𝑣 = √𝐺𝑀

𝑟), where G is the gravitational constant.

(3 marks)

(b) Explain the advantage of launching a low-altitude equatorial-orbit satellite in a west-to-east

direction.

(2 marks)

𝐅 =𝐆𝐦𝐌


𝐫

𝐆𝐌

𝐫

𝐯𝟐


𝐯 = √𝐆𝐌

𝐫

This type of launch takes advantage of the Earth’s pre-existing rotational

velocity about its axis which occurs in a west to easterly direction.

The satellite will not need to gain any additional velocity to have an equatorial

orbit if launched from west to east.

This saves money which would be wasted on rocket fuel which would have to be

used to gain the additional velocity (accelerate) should you be foolish enough to

launch east to west.


Two satellites, A and B, orbit the Earth, as shown in the diagram below. Both satellites are in circular orbits.

The radius of satellite B is greater than the radius of satellite A.

(a) On the diagram above, draw and label vectors to represent the acceleration of satellite A and

satellite B.

(2 marks)

(b) Satellite A orbits at a radius of 2.112 x 107 m. Satellite B orbits at a radius of 4.224 x 107m and at a

speed of 3072 ms-1.

(i) Show that the speed v of a satellite moving in an orbit of radius r round a planet of mass M

is given by, (𝑣 = √𝐺𝑀

𝑟), where G is the gravitational constant.

(3 marks)

𝐅 =𝐆𝐦𝐌


𝐫

𝐆𝐌

𝐫

𝐯𝟐


𝐯 = √𝐆𝐌

𝐫

(ii) Hence show that the mass of the Earth is approximately 5.98 x 1024 kg.

(2 marks)

(iii) Calculate the orbital speed of satellite A.

(2 marks)

(iv) Calculate the orbital period of satellite B.

(2 marks)

(v) Geostationary satellites move in an equatorial orbit in the same direction as the Earth's

rotation. Explain why geostationary satellites have orbits of relatively large radius.

(2 marks)

𝐯 = √𝐆𝐌

𝐫 rearranged is 𝑴 =

𝒗𝟐𝒓

𝑮

𝑴 = (𝟑𝟎𝟕𝟐)𝟐𝒙 𝟒. 𝟐𝟐𝟒 𝒙 𝟏𝟎𝟕

𝟔. 𝟔𝟕 𝒙 𝟏𝟎−𝟏𝟏

𝑴 = 𝟓. 𝟗𝟕𝟔 𝒙 𝟏𝟎𝟐𝟒 𝒌𝒈

𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝑴 = 𝟓. 𝟗𝟖 𝒙 𝟏𝟎𝟐𝟒 𝒌𝒈

𝐯 = √𝐆𝐌

𝐫

𝐯 = √𝟔. 𝟔𝟕 𝒙 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟗𝟕𝟔 𝒙 𝟏𝟎𝟐𝟒

𝟐. 𝟏𝟏𝟐 𝐱 𝟏𝟎𝟕

𝐯 = 𝟒. 𝟑𝟑𝟒 𝐱 𝟏𝟎𝟑 𝒎𝒔−𝟏

Unlike planets which have elliptical orbits, human-made satellites have circular

orbits and therefore circular motion formulae may be used:

𝑻 = 𝟐𝝅𝒓

𝒗

𝑻 = 𝟐𝝅(𝟒.𝟐𝟐𝟒 𝒙 𝟏𝟎𝟕)

𝟑𝟎𝟕𝟐 = 8.639x 104 seconds

Geostationary orbits remain above the same point on the Equator and so must

have a period of revolution of 24 hours. The radius of the orbit must be chosen

such that the required centripetal force is exactly supplied by gravity, which

results in a relatively large orbit (35,900 km).

(vi) Two satellites of equal mass orbit the Earth. One satellite has a radius of rx and the other has

a radius of r2x (double the radius of rx). Calculate the ratio Fx : F2x of the gravitational forces

acting on the satellites.

(3 marks)


Astronaut A is on the surface of a moon of radius r. Astronaut B is at a distance of 3r from the centre of the

moon, as shown in the diagram below:

Astronaut A and astronaut B have identical masses. The magnitude of the gravitational force between the

moon and astronaut A is 195 N.

(a) Calculate, using proportionality, the magnitude of the gravitational force between the moon and

astronaut B.

(3 marks)

Doubling the radius of the orbit results in 𝟏

𝟒 the gravitational force due to

the inverse square law.

F1 : F2 = 1 : 4

𝐅 = 𝟏

𝐫𝟐

𝐅𝐁 = 𝐅𝐀

𝟑𝟐

𝐅𝐁 = 𝟏𝟗𝟓

𝟗

𝐅𝐁 = 𝟐𝟏. 𝟕 𝐍

(b) A satellite is in a circular polar orbit around the Earth at an altitude of 8.54 x 105 m.

The mass of the Earth is 5.97 x 1024 kg and its mean radius is 6.38 x 106m.

(i) Calculate the orbital speed of the satellite.

(3 marks)

(ii) State two reasons why low-altitude polar orbits are used in meteorology and surveillance.

(2 marks)

𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒆𝒂𝒓𝒕𝒉 + 𝒂𝒍𝒕𝒊𝒕𝒖𝒅𝒆 𝒂𝒃𝒐𝒗𝒆 𝑬𝒂𝒓𝒕𝒉’𝒔 𝒔𝒖𝒓𝒇𝒂𝒄𝒆

= 8.54 x 105 + 6.38 x 106 = 7.234 x106 m

𝐯 = √𝐆𝐌

𝐫

𝐯 = √𝟔. 𝟔𝟕 𝒙 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟗𝟕 𝒙 𝟏𝟎𝟐𝟒

𝟕. 𝟐𝟑𝟒 𝐱 𝟏𝟎𝟔

𝐯 = 𝟕. 𝟒𝟐 𝐱 𝟏𝟎𝟑 𝒎𝒔−𝟏

ANY TWO OF:

They have higher resolution or produce clearer images.

They give almost global coverage.

They have short periods so they can see parts of the Earth more than once a day.


Some satellites have geostationary orbits and some satellites have polar orbits

(a) State two differences between geostationary orbits and polar orbits.

(3 marks)

(b) Derive the formula (T2 = 4𝜋2𝑟3

Gm2) given the formulas 𝑣 =

2𝜋𝑟

𝑇 and 𝑣2 =

𝐺𝑚2

𝑟

(3 marks)

(c) Rearrange this formula to show that the period of satellite motion is given by the formula

r = √𝐺𝑚2𝑇2

4𝜋2

3

(2 marks)

A geostationary satellite stays above the same spot on earth whereas a polar one does not

A geostationary satellite orbits above the equator, whereas the polar one goes

approximately over the poles

OR:

A geostationary satellite has a larger radius than a polar one.

A geostationary satellite has a longer period than a polar one.

A polar travels faster than a geostationary one.

𝐯 = 𝟐𝛑𝐫

𝐓 𝐯𝟐 =

𝐆𝐦𝟐

𝐫

𝐯𝟐 = 𝟐𝟐𝛑𝟐𝐫𝟐

𝐓𝟐 𝐯𝟐 = 𝐆𝐦𝟐

𝐫




𝐆𝐦𝟐

𝐓𝟐 = 𝟒𝛑𝟐𝐫𝟑

𝐆𝐦𝟐


𝟒𝛑𝟐


𝟒𝛑𝟐

𝟑

(d) Hence determine the altitude of a satellite in a geostationary orbit around the Earth. The mass of

the Earth is M = 5.97 x 1024kg and its radius is R = 6.4 x 106m.

(4 marks)


The binary star system known as Sirius is shown, at one point in time, in the diagram below.

The mass of Sirius A, measured using data obtained from the Hubble Space Telescope in 2005, is much

larger than that of its partner star, Sirius B.

(a) On the diagram above, draw vectors to show the gravitational force acting on each of these stars at

this point in time.

Equal in length (by eye);

Opposite in direction (towards one another) (2 marks)

(b) Explain why Newton's law of universal gravitation is consistent with Newton's third law of motion.

(3 marks)

T = 24 x 3600 = 𝟖. 𝟔𝟒 𝐱 𝟏𝟎𝟒s


𝟒𝛑𝟐

𝐫𝟑 = 𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝒙 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒 𝐱 (𝟖. 𝟔𝟒 𝐱 𝟏𝟎𝟒)𝟐

𝟒𝛑𝟐

𝐫 = √𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝒙 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒 𝐱 (𝟖. 𝟔𝟒 𝐱 𝟏𝟎𝟒)𝟐

𝟒𝛑𝟐

𝟑

𝐫 = 𝟒. 𝟐𝟐𝟕 𝐱 𝟏𝟎𝟕𝐦

Altitude above Earth = 𝟒. 𝟐𝟐𝟕 𝐱 𝟏𝟎𝟕 - 6.4 x 106 = 3.6 x 107 m

Newton’s 3rd Law states that if Sirius A exerts a force on Sirius B, then Sirius B will exert an equal but

oppositely directed force Sirius B. The gravitational force between two masses is attractive the forces

are oppositely directed and the magnitude of gravitational force as prescribed by Newton's law of

Universal Gravitation, is directly proportional to the product of the objects masses. The two forces will

be the equal magnitudes as:

MA x MB = MB x MA;

Two forces act on different objects.

(c) The polar-orbiting satellite NOAA-N was launched in May 2005, as shown in the photograph below:

The satellite is now moving in a circular orbit above the Earth's surface at an altitude of 870 km.

The mass of the Earth is 5.97 x 1024 kg and its mean radius is 6.38 x 106 m.

(i) Show that the orbital speed of the satellite is 7.41 x103 ms-1

(2 marks)

(ii) Calculate the magnitude of the acceleration due to gravity at the satellite's altitude.

(3 marks)

(iii) Explain why the centre of the circular orbit of any Earth satellite must coincide with the

centre of the Earth.

(3 marks)

r = 6.38 x 106 + 870 x 103 = 7.25 x 106m

𝐯 = √𝐆𝐌

𝐫 = √𝟔.𝟔𝟕 𝒙 𝟏𝟎−𝟏𝟏 𝐱 𝟓.𝟗𝟕 𝒙 𝟏𝟎𝟐𝟒

𝟕.𝟐𝟓 𝐱 𝟏𝟎𝟔

𝐯 = 𝟕. 𝟒𝟏𝟏 𝐱 𝟏𝟎𝟑

𝒎𝒔−𝟏

𝐠 =𝐆𝐦𝟐

𝐫𝟐

𝐠 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒

(𝟕. 𝟐𝟓 𝐱 𝟏𝟎𝟔)𝟐

𝐠 = 𝟕. 𝟓𝟖 𝐦𝐬−𝟐

The gravitational pull of the earth is towards the centre of the Earth. This is the

net force acting on the satellite and so this force must cause centripetal

acceleration. Therefore must point towards the centre of the circular orbit of

the satellite.


A satellite of mass m moves in a circular orbit of radius r about a planet of mass M, as shown in the

diagram below.

(a) Using Newton's law of universal gravitation and Newton's second law of motion, show that the

period T of the satellite is given by (T2 = 4𝜋2𝑟3

Gm2)

(5 marks)

(b) Two identical satellites X and Y, with orbital radii 16r and r respectively, move in circular orbits

about the Earth, as shown in the diagram below.

(i) Using proportionality, calculate the ratio Tx : Ty of the satellites' orbital periods.

(3 marks)

𝐅 =𝐆𝐦𝐌


𝐫

𝐆𝐌

𝐫

𝐯𝟐


𝐯 = √𝐆𝐌

𝐫

𝐯 = 𝟐𝛑𝐫

𝐓 𝐯𝟐 =

𝐆𝐦𝟐

𝐫

𝐯𝟐 = 𝟐𝟐𝛑𝟐𝐫𝟐

𝐓𝟐 𝐯𝟐 = 𝐆𝐦𝟐

𝐫




𝐆𝐦𝟐

𝑰𝒇 , 𝐓𝟐 = 𝟒𝛑𝟐𝐫𝟑

𝐆𝐦𝟐 𝒕𝒉𝒆𝒏 𝐓𝟐 ∝ 𝐫𝟑

𝐓 ∝ √𝐫𝟑

Tx : Ty = √𝐫𝒙𝟑 ∶ 𝐫𝒚

𝟑

Tx : Ty = √𝟏𝟔𝐫𝟑 ∶ 𝐫𝒚𝟑

Tx : Ty = √𝟒𝟎𝟗𝟔 ∶ 𝟏

Tx : Ty = 64 : 1

(ii) Satellite X is a geostationary satellite moving in the Earth's equatorial plane.

Explain why this satellite must move in a particular orbit of relatively large radius.

(3 marks)


The centripetal acceleration of the Earth in its orbit around the Sun has a magnitude of 5.90 × 10−3 ms−2.

This acceleration is caused by the gravitational force that the Sun exerts on the Earth.

The mass of the Earth is 5.97×1024 kg and the mean radius of the Earth’s orbit around the Sun is

1.50 × 1011 m.

(a) Show that the magnitude of the gravitational force that the Sun exerts on the Earth is 3.52 × 1022 N.

(2 marks)

(b) Hence determine the mass of the Sun.

(3 marks)

Since 𝐓𝟐 ∝ 𝐫𝟑 or since T depends on r

As T = 24 hours which is a relatively large period, then R has to be relatively large.

𝐅 = 𝐦𝐚

𝐅 = 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒 𝒙 𝟓. 𝟗𝟎 𝒙 𝟏𝟎−𝟑

𝐅 = 𝟑. 𝟓𝟐 𝐱 𝟏𝟎𝟐𝟐 𝑵


𝐫𝟐

𝐦𝟐 =𝐅𝐫𝟐

𝐆𝐦𝟏

𝐦𝟐 =𝟑. 𝟓𝟐 𝐱 𝟏𝟎𝟐𝟐 𝒙 (𝟏. 𝟓𝟎 𝒙 𝟏𝟎𝟏𝟏)𝟐

𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟓. 𝟔𝟗𝐱 𝟏𝟎𝟐𝟔

𝐦𝟐 = 𝟏. 𝟗𝟗 𝐱 𝟏𝟎𝟑𝟎 𝐍

(c) Two satellites are moving in circular orbits around the Earth, Orbit A and Orbit B, as shown in the

diagram below. The radius of Orbit B is double the radius of Orbit A.

(i) Calculate the ratio Speed of satellite in orbit A

Speed of satellite in orbit B

(3 marks)

(ii) Explain why the centripetal acceleration of a satellite in an orbit of constant radius is independent of the

mass of the satellite.

(2 marks)

𝐯 = √𝐆𝐌

𝐫 ∴ 𝐯 ∝ √

𝟏

𝐫

𝒗𝒂

𝒗𝒃=

√𝟏𝐫𝒂

√𝟏𝐫𝒃

(𝐬𝐢𝐧𝐜𝐞 𝐆 𝐚𝐧𝐝 𝐌 𝐚𝐫𝐞 𝐭𝐡𝐞 𝐬𝐚𝐦𝐞 𝐟𝐨𝐫 𝐛𝐨𝐭𝐡 𝐬𝐚𝐭𝐞𝐥𝐥𝐢𝐭𝐞𝐬 𝐲𝐨𝐮 𝐜𝐚𝐧 𝐜𝐚𝐧𝐜𝐞𝐥 𝐭𝐡𝐞𝐦 𝐨𝐮𝐭)

√𝐫𝒂

√𝐫𝒃 =

√𝐫𝒂

√𝟐𝐫𝒃 (𝐜𝐚𝐧𝐜𝐞𝐥 𝐨𝐮𝐭 𝐫 𝐟𝐫𝐨𝐦 𝐛𝐨𝐭𝐡) = √𝟐

𝐯 = √𝐆𝐌

𝐫

Since the speed of the satellite depends only on the mass of the earth and not the mass of

the satellite, then the centripetal acceleration is also independent of the mass of the satellite.

(d) The diagram below shows two circular paths around the Earth. Path 1 is not a possible satellite orbit. Path 2

is a low-altitude polar orbit.

(i) Explain why Path 1 is not a possible satellite orbit.

(3 marks)

(ii) State one reason why satellites in low-altitude polar orbits are often used for surveillance, and explain

your answer.

(2 marks)

The centre of the orbit in the case of path 1 does not coincide with the centre of

the earth.

The satellite's orbit cannot be stable.

This is because the gravitational force must act towards the centre of the earth

and toward the centre of the orbit in order to provide the centripetal

acceleration for uniform circular motion.

Low-altitude orbits are closer to the surface of the earth. This produces greater

resolution.

This means that images show greater detail and are therefore more useful for

surveillance.

(e) The Iridium satellite network consists of sixty-six communication satellites orbiting the Earth at a radius of

7.18 × 106 m, with a speed of 7.46 × 103 ms−1.

On 10 February 2009 the Iridium 33 satellite collided with the Russian Cosmos 2251 satellite above Siberia.

The damaged Iridium 33 satellite was replaced in the network by a spare satellite that was orbiting at a

lower radius.

Calculate the time that a satellite in the Iridium network takes to complete one orbit of the Earth. Give your

answer to the correct number of significant figures.

(2 marks)


(a) Explain why a geostationary satellite must orbit in a west-to-east direction.

(2 marks)

(b) Explain one advantage of launching equatorial-orbit satellites in a west-to-east direction.

(2 marks)

𝑻 = 𝟐𝝅𝒓

𝑽 (since orbit is approximately circular as it is a human-made satellite)

𝑻 = 𝟐𝝅𝟕. 𝟏𝟖 𝒙 𝟏𝟎𝟔

𝟕. 𝟒𝟔 𝒙 𝟏𝟎𝟑

T = 6.05 x 103 sec

A geostationary satellite is one that remains over a fixed point of the earth's

surface.

This can only occur if it travels in the same direction that the earth rotates

(ie west-to-east).

By launching a satellite in the same direction that the earth spins, the speed of

the earth's spin contributes to the launch speed of the satellite.

This makes the launch more economical as it saves on energy and hence fuel.

The diagram below shows two isolated, spherically symmetric objects. The mass of Object A is much larger

than the mass of Object B.

On the diagram above, draw vectors to show the gravitational forces that these objects exert on each other.

HINT: Despite one mass being larger, the gravitational force acting on each object is equal

in magnitude but opposite in direction.


In November 2012 parts of the world will experience a total solar eclipse. During such an eclipse the Earth,

the Moon, and the Sun are in a straight line. The Moon is between the Earth and the Sun.

In this alignment the distance between the Earth and the Moon is 3.85 x 108 m, and the distance between

the Moon and the Sun is 1.50 x 1011 m.

The mass of the Earth is 5.97 x 1024 kg.

The mass of the Moon is 7.35 x 1022 kg.

The mass of the Sun is 1.99 x 1030 kg.

Determine the magnitude of the ratio force on the moon due to earth

force on moon due to sun

(4 marks)

force on the moon due to earth 𝐅 =𝐆𝐦𝟏𝐦𝟐

𝐫𝟐

𝐅 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟕. 𝟑𝟓 𝐱 𝟏𝟎𝟐𝟐 𝐱 𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒

(𝟑. 𝟖𝟓 𝐱 𝟏𝟎𝟖)𝟐

𝐅 = 𝟐. 𝟎𝟎 𝐱 𝟏𝟎𝟐𝟎 𝑵

force on the moon due to sun 𝐅 =𝐆𝐦𝟏𝐦𝟐

𝐫𝟐

𝐅 =𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏 𝐱 𝟕. 𝟑𝟓 𝐱 𝟏𝟎𝟐𝟐 𝐱 𝟏. 𝟗𝟗 𝐱 𝟏𝟎𝟑𝟎

(𝟏. 𝟓𝟎 𝐱 𝟏𝟎𝟏𝟏)𝟐

𝐅 = 𝟒. 𝟎𝟎 𝐱 𝟏𝟎𝟐𝟎 𝑵

𝐟𝐨𝐫𝐜𝐞 𝐨𝐧 𝐭𝐡𝐞 𝐦𝐨𝐨𝐧 𝐝𝐮𝐞 𝐭𝐨 𝐞𝐚𝐫𝐭𝐡

𝐟𝐨𝐫𝐜𝐞 𝐨𝐧 𝐦𝐨𝐨𝐧 𝐝𝐮𝐞 𝐭𝐨 𝐬𝐮𝐧 =

𝟐.𝟎𝟎 𝐱 𝟏𝟎𝟐𝟎 𝑵

𝟒.𝟎𝟎 𝐱 𝟏𝟎𝟐𝟎 𝑵 = 0.5 So ratio is 1 : 2

The Quick Bird satellite is used to create images of the Earth. One such image is shown below left.

The satellite orbits at an altitude of 482 km, and has a mass of 9.5 x 102 kg.

The International Space Station (shown in the image below right) orbits at an altitude of 390 km, and has a mass of 4.2 x 105 kg.

(a) State whether the QuickBird satellite orbits the Earth at a faster or slower speed than the

International Space Station. Give a reason for your answer.

(2 marks)

(b) State any effect that the different masses of the satellites will have on their speeds. Give a reason

for your answer.

(2 marks)

(c) State one advantage of the QuickBird satellite's low-altitude orbit.

(1 mark)

Low-altitude orbits are closer to the surface of the earth. This produces greater

resolution.

𝐯 = √𝐆𝐌

𝐫

Since the speed of the satellite depends only on the mass of the earth and not the mass of

the satellite, then the centripetal acceleration is also independent of the mass of the

satellite.

Quickbird orbits FASTER. 𝐯 = √𝐆𝐌

𝐫

Since the radius of quickbird is much smaller in magnitude (being closer to the centre of the

Earth), the magnitude of velocity will have to be larger to maintain an orbit as the radius is

inversely proportional to the velocity of the satellite.

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YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONSstage2physicsahs.weebly.com/uploads/4/6/4/2/46424629/gravitation... · than the planet, M. Assume that the satellite moves around the