Question 37: Gravitation

Question

A constant potential energy of a satellite is given as
$$\mathrm{PE}=r(\mathrm{KE})$$
whee, PE = potential energy
and KE = kinetic energy.
The value of $$r$$ will be

Accepted Answer

As we know that, PE of satellite revolving at height $$h$$ from the surface of the earth.
$$ =  - {{G{M_e}m} \over {{R_e} + h}}$$
KE of satellite, $$ = {1 \over 2}{{G{M_e}m} \over {({R_e} + h)}}$$
Therefore, PE = $$-$$2KE
$$	herefore$$ $$r =  - 2$$

Answer

$$-1$$

Answer

$$-2$$

Answer

$$\frac{-1}{2}$$

Answer

$$\frac{-3}{2}$$

Gravitation

A constant potential energy of a satellite is given as $\mathrm{PE}=r(\mathrm{KE}) whee, PE = potential energy and KE = kinetic energy. The value of r$ will be

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