The difference in PE of an object of mass 10kg when it is taken from a height of 6400km to 12800km from the surface of the earth is

Difference in potential energy is given by: PE=-GMm\left(\dfrac {1}{R_1}-\dfrac {1}{R_2}\right) But, R_1 = 6400 km and R_2 = 12800 km \Rightarrow PE=-6.67\times10^{-11}\left(5.97\times10^{24}\right) (10) \left(\dfrac {1}{6400}-\dfrac {1}{12800}\right)\times10^{-3} \Rightarrow PE = 0.00311\times10^{11}J

Gravitation - Problem solving tips

Q: Two lead spheres of same radius are in contact with each other. The gravitational force of attraction between them is F. If two lead spheres of double the previous radius are in contact with each other, the gravitational force of attraction between them will be:

F = \displaystyle \frac{GM_1M_2}{R^2} M = \rho \times \dfrac{4}{3} \pi R^3 \therefore M \propto R^3 \therefore F \propto R^4 \therefore If radius doubles, force becomes 2^4 times i.e. 16 times.

Q: The gravitational force of attraction between two bodies at a certain distance is 10\ N . If the distance between them is doubled, the force of attraction:

Let the two bodies have masses m_1 , m_2 F_G = \dfrac{Gm_1 m_2}{r^2} = 10\ N When distance between them is doubled, F'_G = \dfrac{Gm_1 m_2}{4r^2} = \dfrac{10}{4} \therefore \dfrac{F'_G}{F_G} = 0.25 So, force of attraction decreases by 75\% .

Q: The value of g at a height of 100km from the surface of the Earth is nearly (Radius of the Earth = 6400km) ( g on the surface of the Earth = 9.8m/s^{2} )

Acceleration due to gravity changes with the height from Earth's surface as: g^{'} = g(1-\dfrac {2h}{R}) \Rightarrow g^{'} = 9.8 (1-\dfrac {2\times 100}{6400}) \Rightarrow g^{'} = 9.8(1-\dfrac {1}{32})=9.49 m/s^2

Q: If g on the surface of the Earth is 9.8\ ms^{-2} , its value at a height of 6400 km is: (Radius of the Earth = 6400km )

The expression for g is given by: g_R =\dfrac {GM}{R^2} The value of g changes according to g_h=\dfrac {GM}{(R+h)^2} Therefore, \displaystyle \dfrac{g_{h}}{g_R} = \dfrac{R^{2}}{(R+h)^{2}} Here, h = R So, \dfrac {g_h}{g_R}=(\dfrac {R}{R+h})^2 \Rightarrow \dfrac {g_h}{g_R}=(\dfrac {R}{R+R})^2 \Rightarrow g_h=g_R\times\dfrac {1}{4} = 9.8\times 0.25=2.45\ m/s^2

Q: At a place, the value of 'g' is less by 1% than its value on the surface of the Earth (Radius of Earth, R=6400 km ). The place is :

Let acceleration due to gravity at a place be g_d . If g is acceleration due to gravity at the surface then, g_d=g(1-\frac {d}{R}) 0.99g=g(1-\frac {d}{R}) \frac {d}{R}=(0.01) d=0.01\times 6400=64 km below the surface of earth

Q: If g on the surface of the Earth is 9.8 ms^{-2} , then it's value at a depth of 3200 km (Radius of the earth = 6400 km ) is

The value of gravity changes as we move away from or towards the centre of the Earth.This is given by: g_{R} =\dfrac{GM}{ R^{2} } , where M is the mass of a planet of radius R So, M=\dfrac{4}{3}\pi R^{3}\rho ; substituting in above equation \Rightarrow g_R = G \dfrac{4}{3} \pi \rho \times R Since we want the value of g at depth( d ) from the Earth's surface, we replace R by (R-d) \Rightarrow g_d = G \dfrac{4}{3} \pi \rho \times (R-d) \Rightarrow \dfrac{g_{R}}{ g_{d} } = \dfrac{R}{R-d} \Rightarrow \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{d}{R}) The given depth in the problem is d = 3200 km, substituting we get, \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{3200}{6400}) {g_{d}}=9.8\times(1- \dfrac{3200}{6400}) {g_{d}}=9.8/2 {g_{d}}=4.9 m s^{-2}

Q: For a given density of the planet, the orbital period of a satellite near the surface of planet of radius R is proportional to:

From Kepler's third law, \dfrac{T^2}{r^3}=\dfrac{4\pi^2}{GM} where r is radius, semi-major axis of the elliptical orbital of satellite. Near the planet surface r=R ; M=\cfrac{4}{3} \pi R^3 \rho So, T=\sqrt{\dfrac{3\pi}{G\rho}} \propto R^{0}

Q: The distance between the Sun and the Earth is r and the Earth takes time T to make one complete revolution around the Sun. Assuming the orbit of the Earth around the Sun to be circular, the mass of the Sun will be proportional to

Let m and M be the masses of the earth and the sun respectively and v the speed of the earth in circular orbit. To keep the earth in circular orbit, the gravitational force \dfrac {GMm}{r^2} must balance the centripetal force \dfrac {mv^2}{r} , i.e. \dfrac {GMm}{r^2}=\dfrac {mv^2}{r} or M=\dfrac {v^2r}{G} Also, v=\dfrac {2\pi r}{T} . Using this, we get M=\dfrac {4\pi^2r^3}{T^2G} or M\propto \dfrac {r^3}{T^2} .Hence the correct choice is (c).

Q: If a body of mass m has to be taken from the surface to the earth to a height h = R, then the amount of energy required is (R = radius of the earth)

We know that,gravitational potential energy U=-\dfrac {GM_em}{R} ....(i)and gravitational kinetic energy K=\dfrac {1}{2} \cdot \dfrac {GM_em}{R} ...(ii) \therefore Total energy of a body is E=U+K \dfrac {GM_em}{R}+\frac {GN_em}{2R} \dfrac {2GM_em+GM_em}{2R}=-\frac {GM_em}{2R} But, acceleration due to gravity (g) in terms of gravitational constant (G) is \frac {GM_e}{R^2} ...(iii) \therefore \frac {GM_e}{R^2}\times R^2 \times \frac {me}{2R} [from eq. (iii)] = g\times R^2 \times \frac {m}{2R} (Cancelation of negative sign, because energy can never be negative) =\dfrac {gmR}{2}=\dfrac {mgR}{2}

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Problem solving tips

5 min read

Gravitation

- Want to score better marks in exams? Have a look at some problem solving tips.

Tip 1: Problem related to universal law of gravitation:

The universal law of gravitation includes only one formula i.e. $F = G \frac{m _{1} m _{2}}{r ^{2}}$ .
If the force or the ratio of force between the two conditions is to be determined, then find out the quantities that remains constant in two situations.
While taking the ratio, the constant quantities will be eliminated and the remaining change in the parameters can be addressed in terms of one another to make the solution simple and comprehensive.

Let's practice some problems based on it:

Two lead spheres of same radius are in contact with each other. The gravitational force of attraction between them is F. If two lead spheres of double the previous radius are in contact with each other, the gravitational force of attraction between them will be:

2 F

32 F

8 F

16 F

View solution

The gravitational force of attraction between two bodies at a certain distance is

10 N

. If the distance between them is doubled, the force of attraction:

decreases by

50 %

decreases by

75 %

increases by

50 %

increases by

75 %

View solution

Tip 2: Problem related to calculation of change in value of acceleration due to gravity (due to height).

The value of 'g' with the change in height above the surface of the Earth or with the increasing altitude is given in two different manner.

Thus, it is very necessary to keep in mind if the height is comparable to the radius of Earth or not and then use appropriate formula.

Let's practice some problems based on it:

The value of

g

at a height of

100 k m

from the surface of the Earth is nearly (Radius of the Earth

=

6400km) (

g

on the surface of the Earth

=

9.8 m / s^{2}

)

9.5 m s^{- 2}

8.5 m s^{- 2}

10.5 m s^{- 2}

9.8 m s^{- 2}

View solution

g

on the surface of the Earth is

9.8 m s^{- 2}

, its value at a height of

6400 k m

is: (Radius of the Earth

= 6400 k m

)

4.9 m s^{- 2}

9.8 m s^{- 2}

2.45 m s^{- 2}

19.6 m s^{- 2}

View solution

Tip 3: Problem related to calculation of change in value of acceleration due to gravity (due to depth).

The value of acceleration due to gravity changes due to change in depth from the surface of Earth.
In this cases the value of gravity decreases towards the center using the relation: $g_{d} = g (1 - \frac{d}{R})$
At the center of Earth, the value of 'g' is zero i.e. weightlessness can be experienced at the center of Earth.

Let's practice some problems based on it:

At a place, the value of 'g' is less by 1% than its value on the surface of the Earth (Radius of Earth,

R = 6400 k m

). The place is :

64 km below the surface of the earth

64 km above the surface of the earth

30 km above the surface of the earth

32 km below the surface of the earth

View solution

g

on the surface of the Earth is

9.8

m s^{- 2}

, then it's value at a depth of

3200

k m

(Radius of the earth

= 6400

k m

) is

9.8

m s^{- 2}

z e r o

4.9

m s^{- 2}

2.45

m s^{- 2}

View solution

Tip 4: Problem based on Kepler's third law.

The Kepler's third law gives the relation between the time period and orbital distance of the planet.
We have generally seen the expression of Kepler's third law as: $T^{2} \propto R^{3}$ .
But the elaborated form of the expression with the constants is as: $T^{2} = (\frac{4 π ^{2}}{G M}) R^{3}$
The above expression enables us in determining the mass of a particular body around which a planet (or anything) revolves and further determine other parameters.

Let's practice some problems based on it:

For a given density of the planet, the orbital period of a satellite near the surface of planet of radius

R

is proportional to:

R^{\frac{1}{2}}

R^{\frac{3}{2}}

R^{\frac{- 1}{2}}

R^{0}

View solution

The distance between the Sun and the Earth is

r

and the Earth takes time

T

to make one complete revolution around the Sun. Assuming the orbit of the Earth around the Sun to be circular, the mass of the Sun will be proportional to

\frac{r ^{2}}{T}

\frac{r ^{2}}{T ^{2}}

\frac{r ^{3}}{T ^{2}}

\frac{r ^{3}}{T ^{3}}

View solution

Gravitational Potential Energy Problem Solving Tips

7 mins

Tip 5: Problems based on use of formula for gravitational potential energy.

The use of formula for calculating gravitational potential energy depends on the height or the distance of the body from the surface of the Earth.
The proper use of the formula and determining parameters is shown in the video added below.

Let's practice some problems based on it:

If a body of mass m has to be taken from the surface to the earth to a height h = R, then the amount of energy required is (R = radius of the earth)

m g R

\frac{m g R}{3}

\frac{m g R}{2}

\frac{m g R}{1 2}

\frac{m g R}{9}

View solution

The difference in

P E

of an object of mass

10 k g

when it is taken from a height of

6400 k m

12800 k m

from the surface of the earth is

3.11 \times 1 0^{8}

1.565 \times 1 0^{8}

2.65 \times 1 0^{8}

4.5 \times 1 0^{8}

View solution