8.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

1 View | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Weight of the body, W = 63 N

Acceleration due to gravity at h from the Earth's surface is given by

g' = $\frac{g}{{(\frac{1 + h}{R_{e}})}^{2}}$ where g = acceleration due to gravity on the Earth's surface, $R_{e}$ = Radius of the Earth. h = $\frac{R_{e}}{2}$

g' = $\frac{g}{{(\frac{1 + R_{e} / 2}{R_{e}})}^{2}}$ = $\frac{g}{{(\frac{1 + h}{2 h})}^{2}}$ = (4/9)g

Weight of the body of mass m at a height h is given by

W' = m X g' = (4/9) mg = (4/9) x w = (4/9) x 63 N = 28 N

0 Upvotes 0 Downvotes

Similar Questions for you

In the equation of angular displacement of a particle moving on a circular path is given as is in radian and t in second. Theq angular velocity of the particle at t = 2 s is

alok kumar singh

If an object is having same weight at same distance above and below the surface of earth, find its distance from surface of earth.

alok kumar singh

$\frac{G M m}{{(R + x)}^{2}} = \frac{G M m (R - x)}{R^{3}}$

->R³ = (R + x)² (R – x)

->R³ = (R²– x²) (R + x)

->x² + Rx – R² = 0

$x = \frac{- R \pm \sqrt[]{R + 4 R^{2}}}{2}$

$x = \frac{(\sqrt[]{5 R} - 1)}{2} R$

Two bodies of mass $m$ and $9 m$ are placed at a distance $R$ . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( $G =$ gravitational constant):

alok kumar singh

Stress $= \frac{F}{A} = \frac{T}{A} = \frac{W}{A}$

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : The escape velocities of planet Ad and B are same. But A and B are of unequal mass.

Reason R : The product of their mass and radius must be same. M₁ R₁ = M₂R₂

In the light of the above statements, choose the most appropriate answer from the options given below:

Vishal Baghel

${(v_{e})}_{A} = {(v_{e})}_{B} \Rightarrow \frac{M_{1}}{R_{1}} = \frac{M_{2}}{R_{2}}$

R is not correct.

A solid sphere of radius R gravitationally attracts a particle placed at 3R form its centre with a force F₁,. Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is: