Question

The height of an object from the surface of the earth is 3200 km. If the radius and mass of the earth are 6.4x106m and 6x1024kg, calculate the acceleration due to gravity at that height. Also find the weight of an object of mass 250 kg at that height.

Answer 1

Given:

Height of the object, h= 3200 Km= 3.2×10⁶m

Radius of Earth, $\rm{R_{e}=6.4\times10^{6}m}$

Mass of Earth, $\rm{M_{e}=6\times10^{24}Kg}$

To Find:

1. The acceleration due to gravity of given object at that height
2. Weight of an object of mass 250 kg at that height

Solution:

We know that,

• The value of acceleration due to gravity at height h above the earth's surface is given by

$\pink{\boxed{\bf{g'=\dfrac{g}{\bigg(1+\dfrac{h}{R_{e}}\bigg)^{2}}}}}$

where,

g is acceleration due to gravity at the surface of Earth

Re is the radius of the Earth

Weight= m×g

where,

m is the mass of the object

g is the acceleration due to gravity at that point

Now,

Let the acceleration due to gravity at height 3200 Km be g'

So,

$\longrightarrow\rm{g'=\dfrac{g}{\bigg(1+\dfrac{h}{R_{e}}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{9.8}{\bigg(1+\dfrac{\cancel{3.2\times10^{6}}}{\cancel{6.4\times10^{6}}}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{9.8}{\bigg(1+\dfrac{1}{2}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{9.8}{\bigg(\dfrac{3}{2}\bigg)^{2}}}$

$\longrightarrow\rm{g'=\dfrac{9.8}{\dfrac{9}{4}}}$

$\longrightarrow\rm{g'=\dfrac{39.2}{9}}$

$\longrightarrow\rm{g'=4.35\:m/s^{2}}$

Now,

Let weight of object of mass 250 Kg at height 3200 Km be W

$\longrightarrow\rm{W=m\times g'}$

$\longrightarrow\rm{W=250\times4.35}$

$\longrightarrow\rm{W=1087.5\:N}$