Calculate the force of gravity on a spacecraft in Newton's toward the earth?

Question

Calculate the force of gravity on a spacecraft 25600 km (4 earth radii) above the Earth's surface if its mass is 1300 kg.

Answer in Newton's toward the earth

Ive tried this problem for a few days now and it keeps saying I am doing something wrong. HELP!

Answer 1

Acceleration due to gravity g is inversely proportional to the square of the distance from the centre of the Earth.

Let R = Earth's radius

The spacecraft is 4R height above Earth's surface.

Therefore, spacecraft's distance from the centre of the Earth = R + 4R = 5R

Therefore, g at spacecraft = g at Earth's surface/5^2 = 9.8/25 m/s^2

Force of gravity = mass of spcecraft * g at spacecraft = 1300 * 9.8/25 N

= 52 * 9.8 N = 509.6 N = 5.096 * 10^2 N

Answer 2

1300(9.8)=12740 newtons

Answer 3

The easiest way is to remember how the force of gravity decreases with distance.  Recall that the force is inversely proportional to the square of the distance.

So, let's say that "W" is the force of gravity on the spaceship when it's sitting on the ground.  For gravity calculations, the "distance" is always taken as the distance from the CENTER of the earth.  So when sitting on the surface, the "distance" is 1 earth radius.  So we can write:

(gravity force at 1 earth radius) = W

But we also know that, at the earth's surface, we can write the force of gravity as "mg".  So that means W=mg, and we can write:

(gravity force at 1 earth radius) = mg

Now consider how the force decreases as we go way up.  The problem says, "4 earth radii above the earth's surface."  When you also add in the distance from the surface to the center, that makes it 5 earth radii from the earth's center.  This means the "high" spaceship is 5 times as far away from earth's center as the "low" spaceship.  And, according to the "inverse square" law, that means the gravity force is 1/5², or 1/25, as much as the "surface" gravity.  We can write:

(gravity force at 5 earth radii) = (1/25)(gravity force at 1 earth radius)

= (1/25)(mg)

The problem gives you "m"; and you know that "g" = 9.8m/s².  Just plug the numbers in.

Answer 4

At such an altitude gravity would be reduced by a factor of sixteen. You do the math.