Question

@Abhisar
Planet Z has a rocky surface. At height 3240 km above the surface, the gravitational acceleration is 48 percent of its value at the surface What is the radius of the planet?

7310 km
9120 km
8280 km
6530 km

Answer 1

@aaronq

Answer 2

Let the radius of the planet be R and mass of the planet be M, then gravitational acceleration at a distance d from the surface is \[g = G\frac{M}{(R+d)^2}\]
g at the surface i.e at d = 0 is \[g_0 = G\frac{M}{R^2}\]
Since \(g\) is 48% of \(g_0\) at d = 3240 km
we have \(g = \frac{48}{100} \times g_0\)
\[G\frac{M}{(R+3240)^2} = \frac{48}{100} \times G\frac{M}{R^2}\]
Can you solve this for R ?

Answer 3

There's no m though

Answer 4

Also what's G value

Answer 5

@Abhisar can u help finish this

Answer 6

You don't need G and as they will cancel out !

Answer 7

You have \((R+3240)^2 = \frac{100}{48} \times R^2\)
Can you solve for R now ?

Answer 8

Should it b 48/100 not the reciprocal

Answer 9

It should be 100/48

Answer 10

Can you do the simplification further and solve for R?

Answer 11

I solved and got 3113

Answer 12

That's not an answer though

Answer 13

wait..

Answer 14

It comes out to be \(R = \frac{6400}{13} \times (6 + 5\sqrt3)\) which is around 7217 km. So you can choose first option.

Answer 15

Sorry, *6480, Now you get 7307 km

Answer 16

Thanks