# On the surface of Earth, a spacecraft has a mass of 2.00 * 10^4 kilograms. What is the mass of the spacecraft at a distance of one Earth radius above Earth's surface?

May 29, 2016

Its mass is still $2.00 \cdot {10}^{4}$ kilograms
Its weight is $5.0 \cdot {10}^{3}$ kilograms

#### Explanation:

The mass is unchanged by the position of the spacecraft, but the gravitational pull between the spacecraft and the Earth decreases as the square of the distance between their centres.

One Earth radius above the surface is twice the distance from the centre of the Earth, so the spacecraft will experience $\frac{1}{2} ^ 2 = \frac{1}{4}$ Earth gravity.

Nov 18, 2017

The mass of the spacecraft will always be $2 \cdot {10}^{4}$ kilograms unless it expels mass in some way.

#### Explanation:

The way objects move in a gravitational field is independent of the objects mass. The famous experiment conducted on the Moon shoed that a hammer and a feather fall at the same rate.

If the spacecraft is a rocket then its mass will decrease as is expels fuel as exhaust gas to gain lift.

Assuming that the spacecraft is propelled to an Earth's radius by a rocket then its mass will be almost unchanged. If fact the spacecraft's mass will increase slightly due to its velocity due to relativity.

It is very important not to confuse mass with weight. Mass is the amount of stuff and is measured in kilograms. Weight is the amount of force a mass exerts and is measured in Newtons. Many people give their weight in pounds or kilograms which is incorrect.

Now, Newton's laws describe the effects of gravity accurately enough unless accuracy needs to be high enough to take relativity into account.

Newton's second law relates mass m$to weight, which is a force F#. $F = m g$Where $g = 9.81 m \frac{\setminus}{s} ^ 2$is the acceleration due to gravity at the Earth's surface. Let's round this to 10. So, the weight of the spacecraft on the surface in Newtons is: $F = 2 \cdot {10}^{4} \cdot 10 = 2 \cdot {10}^{5} = 200 , 000 N$Newton also describes the force due to gravity in terms of the distance $r$from the centre of the Earth. $F = \frac{G M m}{r} ^ 2$Where $G$is the gravitational constant and $M$is the mass of the Earth. So, when the spacecraft is at an altitude of an Earth radius, then the distance from the Earth's centre is $2 r$. The the acceleration due to gravity is inversely proportional to the square of the distance. It has one quarter of its weight $5 \cdot {10}^{4} = 50 , 000 N\$, but the same mass.